In triangle ABC the angles satisfy cosA + cosC = sinB. The triangle must be:

Sum to product gives cosA + cosC = 2·cos((A + C)/2)·cos((A − C)/2), and since (A + C)/2 = 90° − B/2 the first factor is sin(B/2). Meanwhile sinB = 2·sin(B/2)·cos(B/2). Cancelling the non-zero sin(B/2) leaves cos((A − C)/2) = cos(B/2), so |A − C| = B, which combined with A + B + C = 180° forces A = 90° or C = 90°. Check with 90°, 50°, 40°: cos90° + cos40° = 0.766 = sin50°. An equilateral triangle fails, since cos60° + cos60° = 1 while sin60° ≈ 0.866.